12.3 The de Rham complex

Let $M$ be an $m$ -dimensional compact orientable manifold with no boundary. By now we are familiar with the de Rham complex,

$$\cdots \stackrel{\mathrm{d}}{\to }{\Omega }^{r-1}(M{)}^{\mathbb{C}}\stackrel{\mathrm{d}}{\to }{\Omega }^r(M{)}^{\mathbb{C}}\stackrel{\mathrm{d}}{\to }{\Omega }^{r+1}(M{)}^{\mathbb{C}}\stackrel{\mathrm{d}}{\to }\dots \text{\hspace{1em}(12.35)}$$

where ${\Omega }^r(M{)}^{\mathbb{C}}=\Gamma (M,{\wedge }^r{T}^{\ast }{M}^{\mathbb{C}})$ . We complexified the forms so that we may apply the AS index theorem. The exterior derivative satisfies ${\mathrm{d}}^2=0$ . To show that (12.35) is an elliptic complex, we have to show that $\mathrm{d}$ is elliptic. To find the symbol for $\mathrm{d}$ , we note that

$$\sigma (\mathrm{d},\xi )\omega =\mathrm{d}(f\tilde{s}){|}_p=\mathrm{d}f\wedge \tilde{s}+f\mathrm{d}\tilde{s}{|}_p=\xi \wedge \omega$$

where $p\in M,\omega \in {\Omega }_p^r(M{)}^{\mathbb{C}},f(p)=0,\mathrm{d}f(p)=\xi ,\tilde{s}\in {\Omega }^r(M{)}^{\mathbb{C}}$ and $\tilde{s}(p)=\omega$ ; see (12.4). We find

$$\sigma (\mathrm{d},\xi )=\xi \wedge .\text{\hspace{1em}(12.36)}$$

This defines a map ${\Omega }^r(M{)}^{\mathbb{C}}\to {\Omega }^{r+1}(M{)}^{\mathbb{C}}$ and is non-singular if $\xi \ne 0$ . Thus, we have proved that $\mathrm{d}:{\Omega }^r(M{)}^{\mathbb{C}}\to {\Omega }^{r+1}(M{)}^{\mathbb{C}}$ is elliptic and, hence, (12.35) is an elliptic complex. Note, however, that the operator $\mathrm{d}:{\Omega }^k(M)\to {\Omega }^{k+1}(M)$ is not Fredholm since $\ker d$ is infinite dimensional. To apply the index theorem to this complex, we have to consider the de Rham cohomology group $H^r(M)$ instead. The operator $\mathrm{d}$ is certainly Fredholm on this space.

Let us find the index theorem for this complex. We note that ${\dim }_{\mathbb{C}}{H}^r(M;\mathbb{C})={\dim }_{\mathbb{R}}{H}^r(M;\mathbb{R})$ . Hence, the analytical index is

$$\begin{array}{rl}\mathrm{ind}\mathrm{d}& =\sum _{r=0}^m(-1{)}^r{\dim }_{\mathbb{C}}{H}^r(M;\mathbb{C})\\ & =\sum (-1{)}^r{\dim }_{\mathbb{R}}{H}^r(M;\mathbb{R})=\chi (M)\end{array}\text{\hspace{1em}(12.37)}$$

where $\chi (M)$ is the Euler characteristic of $M$ . Suppose $M$ is even dimensional, $m=2l$ . The RHS of (12.33) gives the topological index

$$(-1{)}^{l(2l+1)}{\int }_M\mathrm{ch}\left(\underset{r=0}{\overset{m}{⨁}}(-1{)}^r{\wedge }^r{T}^{\ast }{M}^{\mathbb{C}}\right)\frac{\mathrm{Td}\left(T{M}^{\mathbb{C}}\right)}{e(TM)}{|}_{\text{v o l}}.\text{\hspace{1em}(12.38)}$$

The splitting principle yields

$$\begin{array}{rl}& \mathrm{ch}\left(\underset{r=0}{\overset{m}{⨁}}(-1{)}^r{\wedge }^r{T}^{\ast }{M}^{\mathbb{C}}\right)\\ & =1-\mathrm{ch}\left(T^{\ast }{M}^{\mathbb{C}}\right)+\mathrm{ch}\left({\wedge }^2{T}^{\ast }{M}^{\mathbb{C}}\right)+\cdots +(-1{)}^m\mathrm{ch}\left({\wedge }^m{T}^{\ast }{M}^{\mathbb{C}}\right)\\ & =1-\sum _{i=1}^m{\mathrm{e}}^{-x_i}\left(T{M}^{\mathbb{C}}\right)+\sum _{i<j}{\mathrm{e}}^{-x_i}{\mathrm{e}}^{-x_j}\left(T{M}^{\mathbb{C}}\right)+\dots \\ & +(-1{)}^m{\mathrm{e}}^{-x_1}{\mathrm{e}}^{-x_2}\dots {\mathrm{e}}^{-x_m}(T{M}^{\mathbb{C}})\\ & =\prod _{i=1}^m\left(1-{\mathrm{e}}^{-x_i}\right)\left(T{M}^{\mathbb{C}}\right)\end{array}$$

where we have noted that $x_i(T^{\ast }{M}^{\mathbb{C}})=-x_i(T{M}^{\mathbb{C}})$ . [Let $L$ be a complex line bundle and $L^{\ast }$ be its dual bundle. $L\otimes L^{\ast }$ is a bundle whose section is a map $\mathbb{C}\to \mathbb{C}$ at each fibre of $L$ . $L\otimes L^{\ast }$ has a global section which vanishes nowhere (the identity map, for example) from which we can show $L\otimes L^{\ast }$ is a trivial bundle. We have $c_1(L\otimes L^{\ast })=c_1(L)+c_1(L^{\ast })=0$ , hence $x(L^{\ast })=-x(L)$ . The splitting principle yields $x_i(T^{\ast }{M}^{\mathbb{C}})=-x_i(T{M}^{\mathbb{C}})$ .] We also have

$$\mathrm{Td}\left(T{M}^{\mathbb{C}}\right)=\prod _{i=1}^m\frac{x_i}{1-{\mathrm{e}}^{-x_i}}\left(T{M}^{\mathbb{C}}\right)$$

$$e(TM)=\prod _{i=1}^l{x}_i(T{M}^{\mathbb{C}}).$$

Substituting these in (12.38), we have

$$\mathrm{ind}\mathrm{d}={\int }_M(-1{)}^{l(2l+1)}(-1{)}^l\left(\prod _{i=1}^l{x}_i\left(T{M}^{\mathbb{C}}\right)\right)={\int }_{M}e(TM).\text{\hspace{1em}(12.39)}$$

If $m$ is odd, it can be shown that (Shanahan (1978), p22)

$$\mathrm{ind}\mathrm{d}=0\text{\hspace{1em}(12.40)}$$

which is in harmony with the fact that $e(TM)=0$ if $\dim M$ is odd. In any case, the index theorem for the de Rham complex is

$$\chi (M)={\int }_{M}e(TM).\text{\hspace{1em}(12.41)}$$

Example 12.3

Let $M$ be a two-dimensional orientable manifold without boundary. Equation (12.41) reads

$$\chi (M)=\frac{1}{4\pi }{\int }_M{ϵ}^{\alpha \beta }{\mathcal{R}}_{\alpha \beta }=\frac{1}{2\pi }{\int }_M{\mathcal{R}}_{12}(12.42\mathrm{a})$$

which is the celebrated Gauss-Bonnet theorem. For $\dim M=4$ , it reads as

$$\chi (M)=\frac{1}{32{\pi }^2}{\int }_M{ϵ}^{\alpha \beta \gamma \delta }{\mathcal{R}}_{\alpha \beta }\wedge {\mathcal{R}}_{\gamma \delta }.\text{\hspace{1em}(12.42b)}$$

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